 Thank you very much Alina. That was happening here. Sorry. One moment. Yes, thank you very much Alina. So I will be talking about the optimality of the logarithmic upper bound step. I will make clear what I mean by that. And this is, this was a joint project with a research team. We got together in Oaxaca in Mexico. And the team was composed of, well, the surviving members of the team are myself, Emmanuel Carnado, Andres Chirre, and Julian Mejia Cordero. All right. So what are we talking about? Can everybody read my handwriting? It's not worse than on a board. Okay. So an upper bound sieve of Selberg type, or not to be so personalistic, a quadratic upper bound sieve. For me, it's just a choice of a function from the natural numbers to the reals with rho of D equals one for D in a certain initial interval, and rho of D equals zero for D a little bit larger. So you have this function is one at first, some sort of thing, D1, D2. Here it would be zero. And here anything can happen. Here you're free to choose it however you want. All right. And it could be that, so D1 and D2 are given to us as parameters. So we could have to deal with the case of D1 equals one. It's one of the most common ones. It's called the one parameter case, logically enough. All right. So, well, it's a choice of function, but what's the point of choosing that function? The point is to make the following sum s rho, which is the sum for all positive integers up to big n, sum of the following squared. It's the square of the sum over the divisor, z of n of mu of D rho of D, where mu, of course, is the previous function. And the point is to make these as small as possible, given D1 and D2. Okay. And here I should explain why we care. Sorry, a little. It's up here. Okay. Much better. See, I'm having a little bit of a little bit of a tiny upward inconvenience. Okay. Very good. All right. So why do we care? Let me talk then a tiny bit about applications. So there's a classical use of sieves. That's when you are just trying to estimate the number of integers excluded from certain that do not fall. So you have, you're talking about the integers that are forbidden from falling into certain congruence classes, model of p for many p. So congruence class we may forbid is we could forbid the congruence class zero, say. That is, you could be counting a number of integers that are not divisible by any of many prime numbers. Actually stated things. So it's the, it is actually the class zero that we are excluding. I could make things a bit more general, but so numbers because what happens here? You see when, yeah, so evidently, this square being a square is always on zero. And whenever you have that n is a prime number bigger than the two, you're going to have that it has, yes, it, it cannot have any prime device or it's other than itself. So you cannot have any prime device or smaller than the two. It cannot have any device or smaller than big two is except for one. So the only term here is going to be one. And so whenever you have a prime n that n is prime, this inner sum is one. It's a prime bigger than the two. And in general, it's going to be at the zero. So this is an upper bound on the number of prime numbers. That's a simplest case. And you could state more generally. So a classical early application was an upper bound on the number of twin primes. Of course, we believe that there are infinitely many prime numbers. We don't actually know that. But what already Brun managed to show by a different sieve in the early 20th century was that there cannot be so many. I mean, the twin primes must be fairly thin on the ground. So this will be, and that's a very classical application. It's an upper bound on the number again, such that n plus two is also prime. This is old hat for many of you. If not, it doesn't matter. But let me tell you, before I really get started on the main part of my talk, let me tell you about some interesting recent applications of this kind of sieve. By recent, I mean things that have been going on for the last 20 years and that are really an active field of research. So there's what people call enveloping sieves. It's an old name. It's not a perfect name because it's really the, it's not that there is a specific sieve called an enveloping sieve. It's rather that sieves can be used in the mode of enveloping sieves, so to speak. That's what you're not using this sum here, just as an upper bound or to count. Rather, this is used as a weight. So sieves have well-known limitations, so called parity problem that sieves have when used on their own. But then you may want to combine sieves with other powerful techniques. What you do then is you use this weight, which is a very beautiful weight, especially when D2 is not too large. It's distributed. I mean, you can easily prove that it's well distributed to a greater extent than primes themselves are or to a greater extent than the generalized women hypothesis tells you that primes are. So you use it as a weight to buy ascent towards primes and then you work further using other techniques. So you find this, this used to be known to the specialist, some specialist, so you find this in the works of Julian Ramirez. And then the wider number theoretical public, I think, really became aware of this use of the sort of technique with the work of Colson Vincent Gilderim. They can further buy polymaths, main art, and so forth. I'm not being mean to Jean. It's just that Jean's innovation is more or less orthogonal to this. And we can also find this in Greenteau and so forth. Yeah. Personally, I came into the subject in a somewhat different way. It's also that sieves like these can appear in a way that is completely uninvited, so to speak. So you're, you know, you're a happy person using bones identity in your daily life. And then you cause a Schwartz because that's what we do have of the time. And then you are dealt, you have to bound some slight things. And you can do that, though, you know, this is not as big as it looks. This is all size about a constant sense. But then you can ask yourself, well, what if I'm a bit clever and I use a smooth or somebody tells you why, but why are you not smoothing your sums? Always smooth. So you use the most wide once identity and hope that improvements use a smooth once identity and she Schwartz and you are dealt, you are led to some of these four. So row will be some smooth weight sort of seeing that cropped up in my work on three primes, especially in the second version. And also this, this has cropped up in other ways. So see also, for instance, Graham's work on Linux, Linux constant. So what I'm saying is that the quadratic family of seeds is particularly interesting because it's, it's actually quite useful for what when you give certain not completely traditional users to see and it's also naturally enough that it crops up on its own naturally while you are doing other things. It's a sum of squares that does show up spontaneously. All right. Well, let's go back to the problem of making S row as small as possible. So let's look at the, at the case of one parameter first, the one equals one. Then we actually know exactly what is a row for which M row. No, sorry, I'm getting a bit ahead of myself because I defined S row, not M row. Where is my page one? One, second, not a page one. All right. So before I go to D one or M row, let me, so we were at this stage of minimizing. There's a further reason why we care about this particular kind of sieve, namely sieves are usually limited. I mean, if your function, you can define other sieves that are not quadratic in terms of functions. Then you have the sieve support, which in this case goes from one to D two. And if you're, if the square of your seed, this of your, of the length of your sieve support, D two, is greater than began, then you're usually stuck. Your error term blows up. There's nothing you can do. That's not the case for this kind of sieve, at least not necessarily. But let us focus for now. So the simple, it's very simple to see that you can have this expression where M, you can gain a power of love here, usually D one, D two. So we want to minimize M row also possible. And this is important to give that bounce that are of the same order or greater than N. And that's, that's actually important. Something important about quadratic sieves. Then you don't have a reduction to M row. You have a reduction to something that is similar to M row. But for now, let us focus on M row. Sometimes your, the sound is not very good. It's maybe you touch the microphone or the microphone. Okay. Let me see. Because sometimes it goes that the sound goes a bit soft. Now it's okay. Now it's okay to sound. Maybe that I'm getting away from the microphone. Let me raise the microphone perhaps so that the distance of the microphone is constant. Okay. Thank you so much. Yes. I think that might be better. Is this good now or not? It's good. Yes. Thank you. All right. So we want to minimize M row. Very good. So let's consider the one parameter case first 81 equals one. Then M row is minimal. So it's like this. That's exactly this. It's gamma plus So far so good. Well, this is even negative. And that's very nice. So you might think end of story, at least for the one parameter case. But for many applications, they are taking a reason why row celibate, which actually I haven't written down, is not so good because it is, it is not anything continuous or monotonic. It jumps all over the place. You can write, yeah, asymptotically, it's a product of a continuous factor and an arithmetical factor. You can always express it in that way. And by arithmetical I mean that row of n always depends on the visibility properties of n. So it's not going to have any sort of nice plot. And for any, from any purposes, it is better to work with row monotonic of a given continuous age. So it's basically scaling. It's naturally now from this context. So say row of t, row h of t, we're going to do it in this way. Load it to t, load it, load it to 1, t1. Simply so that, I mean, here, this is just a logarithmic rescaling. And here we're bringing it to the same, to the right scale. But do I mean by the right scale? So h is going to be a function on the reals with h of x equals zero. So what you have is that when t is less than d1, this one, this argument is going to be bigger than one. And so you have h, that h of x has to be one. That's what we wanted for t less than d1. This thing is one. For t greater than d2, we're going to have that this is negative. And so h of that will be z again as we want. So row selberg is not of this form. There is one particular choice of h that has been started plenty. So it's the choice, which is just, you know, so it's going to be zero here, it's going to be one here, and it's the simplest choice. It's just a linear function here. Zero x1, zero for x less than zero, one for x greater than one. All right, so this was studied first by Barban and Behoff. I have some names I think on the first slide. Important, they are the first ones to really study this in detail. Motohashi, Graham, the same time period. So the basic theory was really developed in the late 60s to early 80s. And then the subject seems to have plain moral assortment. And what you have later is much later. It kind of suffocates retaken as a spectacular application. It also includes yielding. It's falling close on the first paper. Again, pincis contribution was, yeah, actually does have more to do with these and junks, but polymath, polymath, may not, there's also recent paper by Yvette Wanny. All right, but what you have here is really, as I said, generalizations and some very beautiful applications. Yeah, you also find it in the work of Ian Tao, as I mentioned before. But I would say our feeling was that, yeah, the study of the basic theory had been left partly incomplete. It had been cultivated in the 70s, up to the early 80s. And then there were some gaps there that had, you know, and the recent work had not really had not been its task to fill those gaps and really complete the theory. So what was the state of the theory? The one parameter case, the main term is the same as in Selberg. So the, the, yeah, so since Selberg cv is optimal, then this has to be the optimal main term. So it's a thesis of one of my students. So figure out the second term is 0.6071. Pricky to give a close expression, right, but he managed to show that it is the actual value or that it lies between the, this first five digits are correct. D1 and D2, the main term is 1 over log D2, D1, and it is optimal. Yeah, whether it was known that it's the main, let's say that it was, it was known that the main term was optimal, though it's a little bit hard to, to pinpoint exactly where that's stated. And what I had shown, say this came up while preparing the second version of the three prime proof. So I had worked out, you know, the second order term for the case that we will not be discussing today so much, the case of D2 greater than square root of n. Then the analysis gets trickier in some ways, but in the end you reduce things to expressions that are not the same but are analogous to the ones we will consider. And here, again, the second order term is negative. So I'll tell you in a moment what the second order term here is at any rate. So what about the order term? So this has been, as I said, partially worked out very recently. So the lower order terms were not known and they were partly worked out in, you know, one of the iterations of the infinite three prime manuscript, mostly available on my website by now, end of advertisement, and end of apology for taking so long at any rate. What about lower order terms? Those had been, those were partly worked out but only very recently, they were not known in the 70s and 80s and I haven't seen them anywhere. And also the question is what, so are they optimal, you know, what are the optimal values and when do you get the optimal values? All right, so let me state the main theorem per take. So the one parameter case would be for D1 equals 1, the quantity we want to minimize is, you know, m rho. So let's do it from rho H0. So this confirms Sebastian's results. We are following different methods. So he was following the same kind of methods. So these two results here on the left side, the results from just a year ago or a couple of years ago were following a real analytic approach. We follow a more complex analytic approach and we get the same. In fact, one can do quite a bit better than log cube two here. Let me see the actual truth. Something that we can, and then we don't really have a hard limit for when the computation has to end. And what is really important, is that this is optimal. Not just the main term, but the second order term. So the simple choice that they have made is actually the best choice and not just when it comes to the main term. So you can do better. At least the second order term is so optimal. If there's any wiggle room, it would have to be so far down that I'm skeptical that even that is possible, but we can show that for general, for H continuous, and for you cannot do better than this. What about cell receive? Well, there is one condition. So you need, seems to be scaling, so for fixed age, of course, you could say, well, I can play tricks with that. Yeah, but the condition is just that for this, you have the condition that the total variation of the derivative of age we sign it, and that it be bounded, you know, the constant here will depend on the total variation of each kind. Yes. We have a question from the audience, from Fabian Pazzucci. Fabian, maybe just unmute your microphone and ask away. Hi, just a quick question. You say that the result is optimal for this value of kappa, but not that it's optimal for the value of kappa, the result is optimal period, that the kappa here is optimal. Yes. So it means that the kappa, you have a closed formula for it. Can you define kappa? Yes. Okay, let me go to the next page now. One moment. One moment. It's an integral. It's not that close, but you asked for it, so don't complain when they write a big formula. I'm not scared. Well, I haven't written down what age of t is yet, so this is what, and by the way, the computation here is rigorous, with integral arithmetic, the integration is rigorous, everything is rigorous. And age of t here is an infinite product. At the end, I will talk, if there is time, about how on earth you do rigorous computations with integrals of infinite products, but it is very much possible, and it doesn't have to take a long time. You asked for it, so don't blame me. Thank you. Okay, now it's going to be 1 over log d2, d1, d1. And this is optimal. Again, here it implies constant the total variation of, and to be really technical, we can't do better than, so the main term is optimal, that was known. The second order term is optimal. Hypothetically, there might be improvements in really lower order terms. And how far down it goes, it's a little bit worse in the first case, but there you have it. Okay. And there are several directions in which this may still be taken. There's work to do. Yes, so for instance, the case of this d2 squared is greater than n. I may do that in the final, finally final version of the three-prime book, bear with me, at least to some extent. And with fully explicit constants, let's not talk about that. So there's this thing, it's an interesting question. There are other things that one might consider. So for instances, how do you combine this with a preliminary sieve in my work and the work of Sebastian? So say sieve-only, odd numbers, and so forth. So actually, this second task is quite easy, and we might include it in the papers, I think we will not make it over long. But for instance, if you just are sieving, if you try to, yeah, a note for the real insiders. So you see that the second order term here is negative, and that's really very convenient. For many practical applications, it's not quite large enough to give you Brunt-Each-March without an error term right away. However, if you decide to restrict to odd numbers and you do get Brunt-Each-March, I think Montgomery von Strengths Brunt-Each-March right away. Just trying, other people can listen again. I was just trying to give the real insiders the accurate measure of exactly how strong the result is. All right, very good. So let me get started on the proof. I have 20 minutes or 15 minutes? 15, 20 minutes you can take. This is the advantage of this online seminar that you can go. Yes. So the first half of the work is proving the following proposition. And I should say that I believe this was half known. So not with, the main term was known. And you define rho as before. M rho, which is a quantity we want to minimize, is going to be h prime squared log d2d1. The main term is going to be this. It's going to be dL2 norm squared of h prime. And here, as I was saying, the main term here was known. You can find that in C polymath 14. Of course, sometimes you have to say that something is first one in the literature in a certain place, but it was sort of recent folklore was in the air. For this polymath, you don't even have to make the distinction because polymath is sort of the aggregation of wisdom, common wisdom, to some extent, common wisdom at a certain point. So I could add recent folklore. Some polymath people have told me that they already knew this before starting the project, but I think that's known, that's true by definition for polymath. But it's not to be found in the literature from the 70s. But what we find in polymath is this main term times one plus a little of one. And for our work, we really need to figure out a smaller answer here. And you can see, well, in fact, what we need and prove is also a bilinear version. That right here simply for simplicity, it's proved in exactly the same way. Yeah. Well, once you have that, then it's clear right away that the main term here is going to be, the main term is going to be optimal for H was H0 because it's just going to be Cauchy-Schwarz, right? So it's clear that you're not going to get a better term, a better main term than H0. And then you can work further what happens when you wiggle things a bit. That's why you need a bilinear version of optimality. So you see where this is going. But first, let me sketch for you a proof of the proposition. The method has precedence in the work of Motohashi. And you can find something a bit like that in polymath as well. But one has to be careful with the procedure because we are aiming at a more or less precise error term. And then we will have to iterate to make this smaller. But one thing at a time. So this can be expressed as a double contour integral a Dirichlet series, but you can put some Zetas into it. Then everything converges more nicely. And that's something that will be useful in the future. This really starts as a Dirichlet series where it's not obvious that there are Zetas in there. But once you factor out some Zetas, it becomes something nicer. So here, f of s is the estimating transform. e s 1 s 2 is going to be a product. And this is going to be bounded on a strip around real particle 0. p 1 plus s 1. You don't have to read all of this. The important thing, as I said, is that it's bounded on a strip. All the nastiness goes into the factors of zeta. All right. So forget about that expression for g, all that we need. So g is greater than, say, one-fifth minus one-fifth. So what you start is, so you let's see be your classical contour. So just your classical contour for the prime numbers here, m is placed by 1. And what you do is you move your, so you start with this is you start with two contours to the right of the y-axis. Then you move one of the contours to the classical contour. And you invert the order of summation. Now you shift to the left. And then for, then you're going to pick up a pole, a simple pole at s 2 minus s 1. So you are going to end up with a double integral as your error term. And as your main term, you're going to have the contributions. So for each, the, instead of the integral here, you're just going to have one term, one contribution coming from the simple pole. So you're going to have an integral of contributions. And then what you can do is this integral, which is the main term. Well, it's in the classical contour, but you can displace this classical contour again. Why not? Let it be on the y-axis. This is an integral on the classical contour twice. This is the error term. So what I'm going to have is I displace it to the y-axis. And then let's have this integral. How do you cope with this sort of thing? So by planche del, you actually can work out what will turn out to be the main term of this main term. Because if you just have t squared, FIT squared here, sort of sensible approximation, as we shall see. This is just going to be H prime X squared dx. And so it remains to bound, EIT minus IT. And this converges nicely. This is much nicer than what we had here. So this is O of t to the fourth. As t goes to zero, it's not hard to show. And also you have that this term here, which looks nasty, actually increases very slowly. Are you forgetting anything? No, I think it's correct. And for continuous H, this has faster decay than t squared. So this is all good. This is a basic idea in the one parameter case. That's how you work out what things are for H0, at least at first. At least as far as the main term is concerned. And then for the second order term, what do you do? So for d1, d2, general, this is actually very straightforward. What you have is, must give a better bound for here, with myself, where H, which is the H from the beginning, is dIT minus IT. And F of H is the main transform of H0. So let us focus on H0 right now. That's what we made, since we already know that whatever gives you the optimal main term will have to be H0 plus something smaller that we know from already from the main term that we worked out. Yeah. For H0, this is going to be worked out like this. Yeah. And then you can actually get again from the fact that this is oscillatory. Very good. So just do the analysis. You have some contours and you're happy. For d1 equals 1, you have to do more careful analysis. Some more careful analysis you have that the error term, again, is something that you can do exactly for H equals H0. And then what you do is, you use your previous work to say for a general H, in order to get to help to get the right main term, I'm going to have H0 plus a small displacement and then you recur. All right. I don't want to really get into the details of the recursion in order to make the error term smaller, the iteration really. Let me just tell you something that some of you may be asking yourselves. I already mentioned it once. How to compute this kappa here rigorously. And this is important. It's not just that I keyed something into the computer. We actually prove using a computer that this value has, you know, that it's very six significant digits are such and such. All right. So the syndrome, of course, has different chunks. So in the range C0 epsilon, do a Taylor expansion, rate of T. In the range, T to the infinity, which would use explicit bounds for one over the zeta. And in fact, by use, well, we had to prove them because what happens is complex analytic methods were not being used for cancellation very much because I mean, until recently, there were basically no explicit bounds in the literature for these. And it was believed that they were going to be horrible. Now, so there is there are some bounds by tradition, but they have a constant in front, which is like, best of cases, something like, well, it's more than 1000 has four digits. So that's too much for computations. You don't something that is a million times log is not good enough. So we actually managed to improve traditions bounds using some other bounds by tradition and basically his method, but there were some optimizations with that. And so this is something reasonable. So yeah, um, yeah, and you can use this is probably about 10 to 20 times worse than the reality. But it's already good enough that you can show that when big T say 20,000, the tail of our internal is going to be quite small. And then what remains of course is to, well, how do you deal with H of T, which is an infinite product? Nowadays, yeah, there are finally, you know, consumer and products that, so to speak, that let you do internal arithmetic and rigorous integration without really getting into the guts of things. So we use the internal arithmetic or rather ball arithmetic because we were using our, which has a nice integration feature. The really the question is how to, how do you base approximate? Because what happens is H is slowly convergent product. So how do you approximate it well by a finite product? Well, already in order to express H of T, as something that was clearly bounded on a strip, we had to take out factors of Zeta. So what we do is just we take out more factors of Zeta. So this is a very old idea. The earliest reference, I found this in a paper by somebody called Western in the 20s, which says that basically little would give him that idea. Yeah. And I think this idea is, an idea like this is just, well, the logarithm of this idea is using Pari to some, to accelerate convergence of some infinite sums. I don't think Pari, that's the biggest error bounds that I don't know at any rate. So this is an old idea. And what you have is that H of T, for instance, can write H of T as we already were taking out. You want to get values of Zeta that look like it, that get rid of the lower order terms. And so what you will have is a very rapidly convergent series. You can work out more terms. And when I was giving this talk for the first time just two months ago, and Max Glank, Don Seguir was in the audience and he told me that what he would do, so we have, we do well enough with what I am writing. So a few terms and then we have an infinite product that converges quickly. So it can be truncated after a few terms, and then you just estimate the tail. Yeah. So what Don Seguir was telling us is that in fact, he could show that, yeah, you can actually express H of T as an infinite product of values of Zeta. And then truncate that using, yes, just using your standard facts, standard bounds on Zeta. So that's an alternative way of doing it. Yeah, though, you know, yes, it's actually Peter Morecoz also in the audience pointed out, you do have to do a little bit of this. You have to, at least, include the factors two or two and three, or otherwise your infinite product of values of Zeta doesn't converge that nicely. So some combination of the two. Yeah. And that's basically it. So I think my time is up. Yeah, so that's where we have left things. I hope that you have all found something of interest in this talk. But the main moral is that, yes, the function that people were using is the optimal one and not just as far as the main term is concerned, but also as far as lower terms are concerned, lower terms are concerned, and you can work out the second order term precisely. It has a nice constant. And what's really important is that it's negative. And that helps you for many things in your life. Thank you so much.